Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2012 | June | Q#9

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The line L1 has equation 4y + 3 = 2x.

The point A (p, 4) lies on L1.

a.   Find the value of the constant p.

The line L2 passes through the point C (2, 4) and is perpendicular to L1.

b.   Find an equation for L2 giving your answer in the form ax + by + c = 0, where a, b and c are  integers.

The line L1 and the line L2 intersect at the point D.

c.   Find the coordinates of the point D.

d.   Show that the length of CD is .

A point B lies on Land the length of

The point E lies on Lsuch that the length of the line CDE = 3 times the length of CD.

e.   Find the area of the quadrilateral ACBE.



We are required to find x-coordinate of the point A (p, 4).

If a point lies on the curve (or the line), the coordinates of that point satisfy the equation of the curve  (or the line).

We are given that point A lies on the line Lwhich has equation;

4y + 3 = 2x

We substitute coordinates of point A in the equation of the line L1;

4(4) + 3 = 2(p)

16 + 3 = 2p

2p =19


We are required to find equation of line L2.

To find the equation of the line either we need coordinates of the two points on the line (Two-Point  form of Equation of Line) or coordinates of one point on the line and slope of the line (Point-Slope  form of Equation of Line).

We have coordinates of a point C(2,4) on line L2 but not  slope of the line L2.

Next we need slope of the line Lto write its equation. 

We are given that the line L2 is perpendicular to L1.

If two lines are perpendicular (normal) to each other, then product of their slopes  and  is;

Therefore, if we have slope of line L1, we can find slope of the line L2.

We do not have slope of both lines Land L2. However, we are given equation of line L1 as;

Slope-Intercept form of the equation of the line;

Where  is the slope of the line.

Therefore, we rearrange the given equation of the line L1 to write it in slope-intercept form;



With coordinates of a point on the line Las C(2,4) and its slope   at hand, we can write
equation of the line L2.

Point-Slope form of the equation of the line is;


We are given that the line L1 and the line L2 intersect at the point D and we are required to find the  coordinates of point D.

If two lines (or a line and a curve) intersect each other at a point then that point lies on both lines i.e.  coordinates of that point have same values on both lines (or on the line and the curve).  Therefore, we can equate  coordinates of both lines i.e. equate equations of both the lines (or the  line and the curve).

Equation of the line L1 is;

Equation of the line L2 is;

From equation of the line L1, we can substitute expression for  in equation of the line L2;

Single value of y indicates that there is only one intersection point.

With y-coordinate of point of intersection of two lines (or line and the curve) at hand, we can find the  x-coordinate of the point of intersection of two lines (or line and the curve) by substituting value  of y-coordinate of the point of intersection in any of the two equations.

We choose;

Hence coordinates of point D are .


We are required to find the length of CD.

Expression for the distance between two given points  and is:


We have coordinates of  and . Therefore;


We are given that if point B lies on line L1;

We are also given that the point E lies on L2 such that the length of the line CDE = 3 times the  length of CD.

We are required to find the area of quadrilateral ACBE.

It is becomes much convenient if we sketch the quadrilateral ACBE as shown below.

First of all we draw the line L1 passing through point A(9.5,4) and a fictitious point B on the same  line (because we donot have coordinates of point B).

Then we draw the line L2 which is perpendicular to the line Lat point D (taken arbitrarily between  points A and B) where the two lines intersect as given. Here it is quite possible to draw the line Lbeyond either point A or B outside the segment AB but then it would not make a quadrilateral  ACBE.

Next we sketch points C and E on the line L2. Then join points A,B,C and E to make quadrilateral  ACBE.

This is shown the figure below.


We are required to find the area of this quadrilateral ACBE.

If we join points A and B, the quadrilateral ACBE is divided into two triangle ABC and ABE. 


Expression for the area of the triangle is;

Since CE (the Line L2) and AB (the line L1) are perpendicular, CD and De form the heights of the  two triangles.

We are given that

We have found in (d) that , therefore;

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